The coefficient of linear expansion `'alpha`' of a rod of length `2` m varies with the distance x from the end of the rod as ` alpha = alpha_(0) + alpha_(1) "x" ` where `alpha_(0) = 1.76 xx 10^(-5) "^(@)C^(-1)"and" alpha_(1) = 1.2 xx 10^(-6)m^(-1) "^(@)C^(-1)`. The increase in the length of the rod. When heated through `100^(@)C` is :-

Options

A. `2`cm
B. `3.76`mm
C. `1.2`mm
D. None of these

Explanation

To solve the problem of finding the increase in length of the rod when heated through \(100^\circ C\), we will follow these steps:

### Step 1: Understand the Coefficient of Linear Expansion
The coefficient of linear expansion, \(\alpha\), varies with the distance \(x\) from the end of the rod as:
\[ \alpha = \alpha_0 + \alpha_1 x \]
where
\(\alpha_0 = 1.76 \times 10^{-5} \, ^\circ C^{-1}\) and \(\alpha_1 = 1.2 \times 10^{-6} \, m^{-1} \, ^\circ C^{-1}\).

### Step 2: Set Up the Expression for Change in Length
The change in length \(dL\) of a small segment \(dx\) of the rod at position \(x\) when the temperature changes by \(\Delta T\) is given by:
\[ dL = \alpha \cdot L_0 \cdot dx \cdot \Delta T \]
where \(L_0\) is the original length of the segment, which is \(dx\) in this case.

### Step 3: Integrate Over the Length of the Rod
To find the total change in length \(\Delta L\), we need to integrate \(dL\) from \(0\) to \(2\) m:
\[ \Delta L = \int_0^2 (\alpha_0 + \alpha_1 x) \cdot dx \cdot \Delta T \]
Substituting \(\Delta T = 100^\circ C\):
\[ \Delta L = 100 \int_0^2 (\alpha_0 + \alpha_1 x) \, dx \]

### Step 4: Calculate the Integral
Now we will calculate the integral:
\[ \int_0^2 (\alpha_0 + \alpha_1 x) \, dx = \int_0^2 \alpha_0 \, dx + \int_0^2 \alpha_1 x \, dx \]
Calculating each part:
1. \(\int_0^2 \alpha_0 \, dx = \alpha_0 \cdot x \big|_0^2 = \alpha_0 \cdot 2\)
2. \(\int_0^2 \alpha_1 x \, dx = \frac{\alpha_1}{2} x^2 \big|_0^2 = \frac{\alpha_1}{2} \cdot 4 = 2\alpha_1\)

Combining these:
\[ \int_0^2 (\alpha_0 + \alpha_1 x) \, dx = 2\alpha_0 + 2\alpha_1 \]

### Step 5: Substitute Values and Calculate \(\Delta L\)
Now substitute the values of \(\alpha_0\) and \(\alpha_1\):
\[ \Delta L = 100 \cdot (2\alpha_0 + 2\alpha_1) = 100 \cdot 2 \left(1.76 \times 10^{-5} + 1.2 \times 10^{-6} \cdot 2\right) \]
Calculating:
\[ = 100 \cdot 2 \left(1.76 \times 10^{-5} + 2.4 \times 10^{-6}\right) = 100 \cdot 2 \cdot (2.0 \times 10^{-5}) = 100 \cdot 4.0 \times 10^{-5} \]
\[ = 4.0 \times 10^{-3} \, m = 4.0 \, mm \]

### Final Answer
The increase in the length of the rod when heated through \(100^\circ C\) is:
\[ \Delta L = 4.0 \, mm \]